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December 20, 2014

December 20, 2014

Posted by **Anonymous** on Wednesday, November 2, 2011 at 4:17pm.

f(x) = sin^2(x/3)

defined on the interval [ -9.424778, 2.356194].

Rememer that you can enter pi for \pi as part of your answer.

a.) f(x) is concave down on the interval .

b.) A global minimum for this function occurs at .

c.) A local maximum for this function which is not a global

maximum occurs at .

d.) The function is increasing on the region .

- Calculus -
**Steve**, Wednesday, November 2, 2011 at 4:56pmWe went through this already:

Posted by Anonymous on Tuesday, November 1, 2011 at 1:53pm.

f(x) = sin^2(x/2)

defined on the interval [ -5.683185, 1.270796].

Take a look at that one for the method.

- Calculus -
**Anonymous**, Wednesday, November 2, 2011 at 5:11pmi cant see how you figure that out

- Calculus -
**Anonymous**, Wednesday, November 2, 2011 at 6:31pmI cant find the interval in which it concaves down, and that's the only question i need.

- Calculus -
**Steve**, Thursday, November 3, 2011 at 2:16pmf = sin^2(x/3)

f' = 2/3 sin(x/3)cos(x/3) = 1/3 sin(2x/3)

f'' = 2/9 cos(2x/3)

f is concave down when f'' < 0

so, where is cos(2x/3) < 0?

When pi/2 < 2x/3 < 3pi/2

or 3pi/4 < x < 9pi/4

Your interval appears to be about (-3pi/2,pi/2), so you may have to adjust your answer a bit to fit the interval.

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